\(2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(=3^{32}-1< 3^{32}\)

Gợi ý: Sử dụng liên tục tính chất \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

2(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)

= (3 - 1)(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)

= (3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)

= (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)

= (3⁸ - 1)(3⁸ + 1)(3¹⁶ + 1)

= (3¹⁶ - 1)(3¹⁶ + 1)

= 3³² - 1 < 3³² 

TKS CHÂU NHA!

\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

                \(.........\)

\(=\frac{1}{2}\left(3^{32}-1\right)\)\(< \)\(3^{32}-1\)\(=\)\(A\)

Vậy  \(B< A\)

 A=1.853020189*10 \(^{15}\)

B= 9.265100944*10\(^{15}\)

tự so sánh

Xét B ta có:

\(2B=2\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3-1\right)\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3^2-1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3^4-1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3^8-1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(2B=\left(3^{16}-1\right).\left(3^{16}+1\right)\)

\(2B=3^{32}-1\)

\(B=\frac{3^{32}-1}{2}< A=3^{32}-1\)

Vậy B < A

Baì này mình mới làm lúc sáng bạn vào câu hỏi tương tự có đấy

\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=3^{32}-1\)

\(\Rightarrow A=\frac{3^{32}-1}{2}< 3^{32}-1=C\)

Bài 1:

a,\(127^2+146.127+73^2=127^2+2.127.73+73^2\)\(=\left(127+73\right)^2=200^2=40000\)

b,\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(18^8-\left(18^8-1\right)=1\)

\(c,100^2-99^2+98^2-97^2+...+2^2-1\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)\(=199+195+...+3\)

áp dụng công thức Gauss ta đc đáp án là:10100

d, mk khỏi ghi đề dài dòng:

\(\dfrac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}=\dfrac{560000}{40000}=14\)Bài 2:

\(A=\left(2-1\right)\left(2+1\right)\)\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)Cứ tiếp tục ta đc \(A=2^{32}-1< B=2^{32}\)

\(\left(3-1\right)C=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^2+16\right)\)giải như câu a đc:\(\left(3-1\right)C=3^{32}-1\)

\(\Rightarrow C=\dfrac{3^{32}-1}{3-1}=\dfrac{3^{32}-1}{2}< D=3^{32}-1\)

1c,

\(=100^2-99^2+98^2-97^2+...+2^2-1^2\\ =\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\\ =\left(100+99\right)\cdot1+\left(98+97\right)\cdot1+...+\left(2+1\right)\cdot1\\ =100+99+98+97+...+2+1\\ =\dfrac{100\cdot101}{2}=5050\)

Áp dụng liên tục a² - b² = (a - b)(a + b)   để biến đổi . Ta có: 

A = 3³² - 1 = (3¹⁶ - 1)(3¹⁶ + 1) = (3⁸- 1)(3⁸ + 1)(3¹⁶ + 1) = (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 2) = (3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1) =

    = (3 - 1)(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1) = 2.B 

Ta có 2B = \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

2B = (3⁴-1)(3⁴+1)(3⁸+1)(3¹⁶+1)

2B = (3⁸-1)(3⁸+1)(3¹⁶+1)

Tương tự ta đc: 

2B = 3³²-1

B= 3³²-1/2 hay B= A/2

Vậy A>B

ta có : B=(3+1)(3^2  +1)(3^4  +1)(3^8  +1)(3^16  +1)

=>(3-1)B=(3-1)(3+1)(3^2  +1)(3^4 +1)(3^8  +1)(3^16  +1)

=>2B=((3^2  -1)(3^2  +1)(3^4 +1)(3^8 +1)(3^16 +1)

=>2B=(3^4 -1)(3^4 +1)(3^8 +1)(3^16 +1)

=>2B=(3^8 -1)(3^8 +1)(3^16 +1)

=.2B=(3^16 -1)(3^16 +1)

=>2B=3^32  -1

=>B=(3^32  -1)/2, mà A=3^32  -1 nên  B=A/2 hay A=2B

a) \(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)

\(=\left(a^2+\left(-b\right)^2+c^2-2ab+2ac-2bc\right)-\left(b^2-2bc+c^2\right)+2ab-2ac\)

\(=a^2+b^2+c^2-2ab+2ac-2bc-b^2+2bc-c^2+2ab-2ac\)

\(=a^2+b^2-b^2+c^2-c^2-2ab+2ab+2ac-2ac-2bc+2bc\)

\(=a^2\)

Câu 3 kiểm tra lại đề lại với , nếu đúng thì phức tạp lắm, còn sửa lại đề thì là :

\(y^2+2y+4^x-2^{x+1}+2=0\)

\(=>\left(y^2+2y+1\right)+2^{2x}-2^x.2+1=0\)

\(=>\left(y+1\right)^2+\left(\left(2^x\right)^2-2^x.2.1+1^2\right)=0\)

\(=>\left(y+1\right)^2+\left(2^x-1\right)^2=0\)

Dấu = xảy ra khi :

\(\hept{\begin{cases}y+1=0\\2^x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-1\\x=0\end{cases}}}\)

CHÚC BẠN HỌC TỐT........... 

mk chịu

1, Khai triển ra ta được:

\(r\left(x\right)=-\left(9x^2-42x+49\right)+6x-14-17\)

\(=-9x^2+42x-49+6x-14-17\)

\(=-9x^2+48x-80\)

\(=-9x^2+48x-64-16\)

\(=-\left(\left(3x\right)^2-3x.2.8+8^2\right)-16\)

\(=-\left(3x+8\right)^2-16\)

\(Do-\left(3x+8\right)^2\le0\)

\(=>-\left(3x+8\right)^2-16\le-16\)

Dấu bằng xảy ra khi \(3x+8=0=>x=-\frac{8}{3}\)

Vậy giá trị nhỏ nhất là -16 tại \(x=-\frac{8}{3}\)

rút gọn biểu thức

2x

=2x(4x²-4x+1)-3x.(x²-9)-4x(x²+2x+1)

=8x³-8x²+2x-3x³-27x-4x³-8x²-4x

=8x³-16x²-7x³-29x